A clasic mechanics problom with a moving table

[anachin6000]

Homework Statement

"On a horizontal table without friction it's a cube of mass M=70 kg and on the cube it's a body of mass m1=1kg. The body m1 it's conected by a cord with an other body of mass m2=4 kg, situated at hight H=1,5 m above the table. The sistem starts from 0 velocity. What it's the displacement of the cube M by the time when the body m2 touches the table(fig. 1.5.13)?(there is no friction)"

I attached the figure.

Homework Equations

T=m1*a1
T=M*a
G2-T=m2*a2 (G is weight)
a1=a2

The Attempt at a Solution

M*a=m1*a1

h=(a1*t^2)/2
d=(a*t^2)/2
h/d=a1/a=M/m1

d=h*m1/M
But at solutions in the book says d=h*m1/(m1+m2+M).

So, can someone explain me my mistake? And also I'll be very happy if someone can give me a tip for solving the problem using the momentum.

 

[jbriggs444]

Homework Statement

"On a horizontal table without friction it's a cube of mass M=70 kg and on the cube it's a body of mass m1=1kg. The body m1 it's conected by a cord with an other body of mass m2=4 kg, situated at hight H=1,5 m above the table. The sistem starts from 0 velocity. What it's the displacement of the cube M by the time when the body m2 touches the table(fig. 1.5.13)?(there is no friction)"

I attached the figure.

Homework Equations

T=m1*a1
T=M*a
G2-T=m2*a2 (G is weight)
a1=a2

The Attempt at a Solution

M*a=m1*a1

h=(a1*t^2)/2
d=(a*t^2)/2
h/d=a1/a=M/m1

d=h*m1/M
But at solutions in the book says d=h*m1/(m1+m2+M).

So, can someone explain me my mistake? And also I'll be very happy if someone can give me a tip for solving the problem using the momentum.

I would first note that one is apparently supposed to assume that the 4 kg mass (m₂) slides down the side of cube M on massless, frictionless rails and does not sway to the side like a pendulum as cube M is pushed to one side by the force of the pulley.

What is the velocity (in the horizontal direction) of the center of mass of all three masses taken together prior to the start of the problem? What is this velocity after the end? Does this velocity change during the problem?

What does this say about the position of the center of mass during the problem?

 

[anachin6000]

Thank you!

 

[dauto]

Two mistakes.

T=M*a isn't right because the string sways

a1=a2 isn't right either because the pulley is not fixed.

 

[HalfLostAndSearching]

Your mistake is in assuming that the acceleration of the cube (M) is equal to the acceleration of the body (m1). In reality, the acceleration of the cube will be affected by the acceleration of both bodies connected to it (m1 and m2). This is where the momentum approach can be helpful.

Using momentum, we can write:

M*V = m1*v1 + m2*v2

where V is the velocity of the cube and v1 and v2 are the velocities of the bodies m1 and m2 respectively.

Since the system starts from rest, we can also write:

V = v1 = v2 = 0

Solving for V, we get:

V = (m1*v1 + m2*v2)/M

Now, we know that the body m2 will fall from a height of H=1.5m, so its final velocity when it reaches the table will be:

v2 = √(2*g*H) = √(2*9.8*1.5) = 5.42 m/s

Substituting this value in the equation for V, we get:

V = (m1*0 + m2*5.42)/M = (4*5.42)/70 = 0.311 m/s

Now, using the equation for displacement (d = V*t + 1/2*a*t^2), we can find the displacement of the cube M by the time the body m2 touches the table:

d = 0.311*t + 1/2*0*t^2 = 0.311*t

Since we know that the body m2 will reach the table in t seconds, we can substitute this value and find the displacement:

d = 0.311*t = 0.311*√(2*1.5/9.8) = 0.311*0.447 = 0.139 m

So, the displacement of the cube M when the body m2 touches the table will be 0.139 m. This matches with the solution in the book, which takes into account the mass of the cube (M) as well.